Equal Area.pdf

7/29/2019 Equal Area.pdf

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Equal Area Criterion 1.0 Development of equal area criterion As in previous notes, all powers are in per‐unit.

I want to show you the equal area criterion a little

differently than the book does it.

Let’s start from Eq. (2.43) in the book.

aem PPPdt

d H =−=2

2

Re

2 δ

ω (1)

Note in (1) that the book calls ωRe as ωR; however, we

need to use 377 (for a 60 Hz system).

We can

also

write

(1)

as

aem PPPdt

d H =−=

ω

ω Re

2

(2)

Now multiply the left‐hand‐side by ω and the right‐hand

side by dδ/dt (recall ω= dδ/dt) to get:

[ ]dt

d PP

dt

d H em

δ ω ω

ω −=

⎭⎬⎫

⎩⎨⎧2

Re (3)



2

Note:

( )dt

t d

t dt

t d )(

)(2

)( 2ω

ω ω

= (4)

Substitution of (4) into the left‐hand‐side of (3) yields:

[ ]dt

d PP

dt

d H em

δ ω

ω −=

⎭⎬⎫

⎩⎨⎧ 2

Re (5)

Multiply by dt to obtain:

[ ] δ ω ω

d PPd H

em −=2

Re (6)

Now consider a change in the state such that the angle

goes from δ1 to δ2 while the speed goes from ω1 to ω2.

Integrate (6) to obtain:

[ ]∫∫ −=2

1

22

21

2

Re

δ

δ

ω

ω

δ ω ω

d PPd H

em (7)

Note the variable of integration on the left is ω2. This

results in



3

[ ] [ ]∫ −=−2

1

2

1

2

2

Re

δ

δ

δ ω ω ω

d PP H

em (8)

The left‐hand‐side of (8) is proportional to the change in

kinetic energy between the two states, which can be

shown more explicitly by substituting

H=Wk/SB=(1/2)JωR2/SB into (8), for H:

[ ] [ ]∫ −=−2

1

21

22

Re

2

21

δ

δ

δ ω ω ω ω d PP

S J

em

B

R

(8a)

[ ]∫ −=⎥⎦

⎤⎢⎣

⎡−

2

1

2

1

2

2

Re

2

2

1

2

1δ

δ

δ ω ω ω

ω d PP J J

S em

B

R

(8b)

Returning to (8), let ω1 be the speed at the initial

moment of the fault (t=0+, δ=δ1), and ω2 be the speed at

the maximum angle (δ=δr), as shown in Fig. 1 below.

Note that the fact that we identify a maximum angle δ=δr

indicates an implicit assumption that the performance is

stable. Therefore the following development assumes

stable performance.



4

Fig. 1

Since speed is zero at t=0, it remains zero at t=0+. Also,

since δr is the maximum angle, the speed is zero at this

point as well. Therefore, the angle and speed for the two

points of interest to us are (note the dual meaning of δ1:

it is lower variable of integration; it is initial angle):

δ1=δ1 δ=δr

ω1=0 ω2=0

Pm

Pe

δÆδ1 180° 90° δ3

Pm3

Pm1

Pm2

Pe3

Pe1

Pe2

δmδc δr



5

Therefore, (8) becomes:

[ ] [ ]∫ −==−r

d PP H

em

δ

δ

δ ω ω ω

1

02

1

2

2

Re (9a)

We have developed a criterion under the assumption of

stable performance, and that criterion is:

[ ] 0

1

=−∫r

d PP em

δ

δ

δ (9b)

Recalling that Pa=Pm‐Pe, we see that (9b) says that for

stable performance, the integration of the accelerating

power from initial angle to maximum angle must be zero. Recalling again (8b), which indicated the left‐hand‐side

was proportional to the change in the kinetic energy

between the two states, we can say that (9b) indicates

that the accelerating energy must exactly counterbalance

the decelerating energy.

Inspection of Fig. 1 indicates that the integration of (9b)

includes a discontinuity at the moment when the fault is



6

cleared, at angle δ=δc. Therefore we need to break up

the integration of (9b) as follows:

[ ] [ ] 0

1

32 =−+−∫ ∫c r

c

d PPd PP emem

δ

δ

δ

δ

δ δ (10)

Taking the second term to the right‐hand‐side:

[ ] [ ]∫ ∫ −−=−c r

c

d PPd PP emem

δ

δ

δ

δ

δ δ

1

32 (11)

Carrying the negative inside the right integral:

[ ] [ ]∫ ∫ −=−

c r

c

d PPd PP meem

δ

δ

δ

δ

δ δ

1

32 (12)

Observing that these two terms each represent areas on

the power‐angle curve, we see that we have developed

the so‐called equal ‐area criterion for stability. This

criterion says that stable performance requires that the accelerating area be equal to the decelerating area, i.e.,

21 A A = (13)



7

where

[ ]

∫−=

c

d PP A em

δ

δ

δ

1

21

(13a)

[ ]∫ −=r

c

d PP A me

δ

δ

δ 32 (13b)

Figure 2 illustrates.

Fig. 2

Pm

Pe

δÆδ1 180° 90° δ3

Pm3

Pm1

Pm2

Pe3

Pe1

Pe2

δmδc δr

A1

A2



8

Figure 2 indicates a way to identify the maximum swing

angle, δr. Given a particular clearing angle δc, which in

turn fixes A1, the machine angle will continue to increase

until it reaches an angle δr such that A2=A1.

2.0 More severe stability performance Stability performance become more severe, or moves

closer to instability, when A1 increases, or if available A2

decreases. We consider A2 as being bounded from above by δm, because, as we have seen in previous notes, δ

cannot exceed δm because δ>δm results in more

accelerating energy, not more decelerating energy. Thus

we speak of the “available A2” as being the area within

Pe3‐Pm bounded on the left by δc and on the right by δm.

Contributing factors to increasing A1, and/or decreasing

available A2, are summarized in the following four bullets

and corresponding illustrations.



9

1. Pm increases: A1increases, available A2 decreases

Fig. 3

Pm

Pe

δÆδ1 180° 90° δ3

Pm3

Pm1

Pm2

Pe3

Pe1

Pe2

δm

δc δr

A1

A2



10

2. Pe2 decreases: A1 increases.

Fig. 4

Pm

Pe

δÆδ1 180° 90° δ3

Pm3

Pm1

Pm2

Pe3

Pe1

Pe2

δm

δc δr

A1

A2



11

3. tc increases: A1increases, available A2 decreases

Fig. 5

Pm

Pe

δÆδ1 180° 90° δ3

Pm3

Pm1

Pm2

Pe3

Pe1

Pe2

δm

δc δr

A1

A2



12

4. Pe3 decreases: available A2 decreases.

Fig. 6

3.0 Instability and critical clearing angle/time Instability occurs when available A2< A1. This situation is

illustrated in Fig. 7.

Pm

Pe

δδ1 180° 90° δ3

Pm3

Pm1

Pm2

Pe3

Pe1

Pe2

δm

δc δr

A1

A2



13

Fig. 7

Consideration of Fig. 7 raises the following question: Can

we express the maximum clearing angle for marginal

stability, δcr, as a function of Pm and attributes of the

three power angle curves, Pe1, Pe2, and Pe3?

The answer is yes, by applying the equal area criterion

and letting δc=δcr and δr= δm. The situation is illustrated in

Fig. 8.

Pm

Pe

δδ1 180° δ3

Pm3

Pm1

Pm2

Pe3

Pe1

Pe2

δmδc

A1

A2



14

Fig. 8

Applying A1=A2, we have that

[ ] [ ]∫ ∫ −=−cr m

cr

d PPd PP meem

δ

δ

δ

δ

δ δ

1

32 (14)

Pm

Pe

δδ1 180° δ3

Pm3

Pm1

Pm2

Pe3

Pe1

Pe2

δmδcr

A2

A1



15

The approach to solve this is as follows (this is #7 in your

homework #3):

1. Substitute Pe2=PM2sinδ, Pe3=PM3sinδ

2. Do some algebra.

3. Define r1=PM2/PM1, r2=PM3/PM1, which is the same as

r1=X1/X2, r2=X1/X3.

4. Then you obtain:

( )

12

1121

1

coscos

cosr r

r r P

Pmm

M

m

cr −

−+−

=

δ δ δ δ

δ (15)

And this is equation (2.51) in your text.

Your text, section 2.8.2, illustrates application of (15) for

the examples 2.4 and 2.5 (we also worked these

examples in the notes called “ClassicalModel”). We will

do a slightly different example here but using the same

system.



16

Example: Consider the system of examples 2.3‐2.5 in

your text, but assume that the fault is

• At the machine terminalsÎr1=PM2/PM1=0.

• Temporary (no line outage)Îr2=PM3/PM1=1.

The pre‐fault swing equation, given by equation (19) of

the notes called “ClassicalModel,” is

δ δ ω

sin223.28.0)(2Re

−=t H && (16)

with H=5. Since the fault is temporary, the post‐fault

equation is also given by (16) above.

Since the fault is at the machine terminals, then the

fault‐on swing equation has Pe2=0, resulting in:

8.0)(2

Re

=t H

δ ω

&&

(17)

With r1=0 and r2=1, the equation for critical clearing

angle (15) becomes:

( ) mm

M

mcr

P

Pδ δ δ δ coscos 1

1

+−= (18)



17

Recall δm=π‐δ1; substituting into (18) results in

( ) ( )11

1

cos2cos δ π δ π δ −+−= M

m

cr P

P

(17)

Recall the trig identity that cos(π‐x)=‐cos(x). Then (17)

becomes:

( ) 11

1

cos2cos δ δ π δ −−= M

mcr

P

P

(18)

In the specific example of interest here, we can solve for

δ1 from the pre‐fault swing equation, with 0 acceleration,

according to

°==⇒

−=

0925.213681.0

sin223.28.00

1

1

rad δ

δ

(19)

In this case, because the pre‐fault and post‐fault power

angle curves are the same, δm is determined from δ1

according to

°=−=−= 9075.1580925.21180180 1δ δ m (20)

This is illustrated in Fig. 9.



18

Fig. 9

From (16), we see that Pm=0.8 and PM1=2.223, and (18)

can be evaluated as

( )

( )

0674.09330.08656.0

)3681.0cos()3681.0(2223.2

8.0

cos2cos 11

1

−=−=

−−=

−−=

π

δ δ π δ

M

mcr

P

P

Pm

Pe

δδ1 180°

PM1

Pe1

δm δcr



19

Therefore δcr=1.6382rad=93.86°.

It is interesting to note that in this particular case, we can

also express the clearing time corresponding to any clearing angle δc by performing two integrations of the

swing equation.

em PPdt

d H −=

2

2

Re

2 δ

ω (21)

For a fault at the machine terminals, Pe=0, so

mm P H dt

d P

dt

d H

2

2 Re

2

2

2

2

Re

ω δ δ

ω =⇒=

(22)

Thus we see that for the condition of fault at the

machine terminals, the acceleration is a constant. This is

what allows us to successfully obtain t in closed form, as

follows.



20

Integrate (22) from t=0 to t=t:

t P H dt

d

dt P H

dt dt

d

m

t t

t

t

m

t

2

2

Re

0

0

Re

02

2

ω δ

ω δ

=⇒

=

=

=

∫∫ (23)

The left‐hand‐side of (23) is speed. Speed at t=0 is 0.

Therefore

t P H dt

d m

2

Reω δ =

(24)

Now perform another integration of (24):

∫∫ =

t

m

t

t P H

dt dt d

0

Re

02

ω δ

(25)

The left‐hand‐side of (25) is actually an integration with

respect to δ, but we need to change the integration

limits accordingly, where δ(t=0)=δ1, to get

dt t P H

d

t

m

t

∫∫ =0

Re

)(

21

ω δ

δ

δ (26)



21

Performing the integration results in

22)(

2

Re

1

t P

H t

m

ω δ δ =−

(27)

Solving for t yields:

( )1

Re

)(4

δ δ ω

−= t P

H t

m (28)

So we obtain the time t corresponding to any clearing

angle δc, when fault is temporary (no loss of a

component) and fault is at machine terminals, using (28),

by setting δ(t)=δc.

Returning to our example, where we had Pm=0.8, H=5sec,

δ1=0.3681rad, and δcr=1.6382rad=93.86°, we can

compute critical clearing time tcr according to

( ) 2902.03681.06382.1)8.0)(377(

)5(4=−=cr t

The units should be seconds, and we can check this from (28) according to the following:



22

( ) sec)sec)(/(

sec=rad

purad

I have used my Matlab numerical integration tool to test

the above calculation. I have run three cases:

tc=0.28 seconds

tc=0.2902 seconds

tc=0.2903 seconds

Results for angles are shown in Fig. 10, and results for

speeds are shown in Fig. 11.



23

Fig. 11

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-15

-10

-5

0

5

10

15

20

25

tclear=0.2903 seconds

tclear=0.2902 seconds

tclear=0.28 seconds

Time (seconds)

S p e e d

( r a d / s e c )

Fig. 12



24

Some interesting observations can be made for the two

plots in Figs. 11 and 12.

In the plots of angle:

• The plot of asterisks has clearing time 0.2903 seconds

which exceeds the critical clearing time of 0.2902

seconds by just a little. But it is enough; exceeding it by

any amount at all will cause instability, where the rotor

angle increases without bound.

• The plot with clearing time 0.28 second looks almost

sinusoidal, with relatively sharp peaks. In contrast,

notice how the plot with clearing time 0.2902 second

(the critical clearing time) has very rounded peaks. This

is typical: as a case is driven more close to the marginal stability point, the peaks become more rounded.

In the plots of speed:

• The speed increases linearly during the first ~0.28‐0.29

seconds of each plot. This is because the accelerating

power is constant during this time period, i.e., Pa=Pm,

since the fault is at the machine terminals (and

therefore Pe=0).



25

• In the solid plot (clearing time 0.28 seconds), the speed

passes straight through the zero speed axis with a

constant deceleration; in this case, the “turn‐around

point” on the power‐angle curve (where speed goes to

zero) is a point having angle less than δm. But in the

dashed plot (clearing time 0.2902 seconds), the speed

passes through the zero speed axis with decreasing

deceleration; in this case, the “turn‐around point” on

the power angle curve (where speed goes to zero) is a

point having angle equal to δm. This point, where angle

equals δm, is the unstable equilibrium point. You can

perhaps best understand what is happening here if you

think about a pendulum. If it is at rest (at its stable

equilibrium point),

and

you

give

it a push,

it will

swing

upwards. The harder you push it, the closer it gets to

its unstable equilibrium point, and the more slowly it

decelerates as it “turns around.” If you push it just

right, then it will swing right up to the unstable

equilibrium point, hover there for a bit, and then turn

around and come back.



26

• In the speed plot of asterisks, corresponding to

clearing time of 0.2902 seconds, the speed increases,

and then decreases to zero, where it hovers for a bit,

and then goes back positive, i.e., it does not turn‐

around at all. This is equivalent to the situation where

you have pushed the pendulum just a little harder so

that it reaches the unstable equilibrium point, hovers

for a bit, and then falls the other way.

• It is interesting that the speed plot of asterisks

(corresponding to clearing time of 0.2902 seconds)

increases to about 24 rad/sec at about 1.4 second and

then seems to turn around. What is going on here? To

get a better look at this, I have plotted this to 5

seconds, as shown in Fig. 13.



27

Fig. 13

In Fig. 13, we observe that the oscillatory behavior

continues forever, but that oscillatory behavior occurs

about a linearly increasing speed. This oscillatory

behavior may be understood in terms of the power

angle curve, as shown in Fig. 14.



28

Fig. 14

We see that Fig. 14 indicates that the machine does in

fact cycle between a small amount of decelerating

energy and a much larger amount of accelerating

energy, and this causes the oscillatory behavior. The

fact that, each cycle, the accelerating energy is much

larger than the decelerating energy is the reason why

the speed is increasing with time.

Pe

Pm

Decelerating energy

Accelerating energy



29

You can think about this in terms of the pendulum: if

you give it a push so that it “goes over the top,” if

there are no losses, then it will continue to “go round

and round.” In this case, however, the average velocity

would not increase but would be constant. This is

because our analogy of a “one‐push” differs from the

generator case, where the generator is being “pushed”

continuously by the mechanical power into the

machine.

You should realize that Fig. 14 fairly reflects what is

happening in our plot of Fig. 13, i.e., it appropriately

represents our model. However, it differs from what

would actually happen in a synchronous machine. In

reality, once the angle reaches 180 degrees, the rotor magnetic field would be reconfigured with respect to

the stator magnetic field. This is called “slipping a

pole.” Most generators have out‐of ‐step protection

that is able to determine when this happens and would

then trip the machine.



30

4.0 A few additional comments 4.1 Critical clearing time

Critical clearing time, or critical clearing angle, was very

important many years ago when protective relaying was

very slow, and there was great motivation for increasing

relaying speed. Part of that motivation came from the

desire to lower the critical clearing time. Today, however,

we use protection with the fastest clearing times and so there is typically no option to increase relaying times.

Perhaps of most importance, however, is to recognize

that critical clearing time has never been a good

operational performance indicator because clearing time

is not

adjustable

once

a protective

system

is in place.

4.2 Small systems

What we have done applies to a one‐machine‐infinite

bus system. It also applies to a 2‐generator system (see

problem 2.14 in the book). It does not apply to

multimachine systems, except in a conceptual sense.



31

4.2 Multimachine systems

We will see that numerical integration is the only way to

analyze multimachine systems. We will take a brief look at this in the next lecture.

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Equal Area.pdf